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# Higher order differential equations

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### Higher order differential equations

1. 1. Prof. Enrique Mateus NievesPhD in Mathematics Education.1HIGHER ORDER DIFFERENTIAL EQUATIONSHomogeneous linear equations with constant coefficients of order two andhigher.Apply reduction method to determine a solution of the nonhomogeneous equation given in thefollowing exercises. The indicated function y1(x), is a solution of the associated homogeneousequation. Determine a second solution of the homogeneous equation and a particular solutionof the inhomogeneous ED1.x1 ey;yy 224  2. 11  1y;yy3.x1xey;eyyy  3523 4.x1 ey;xyyy  34Problems for group discussion:1. Make a convincing demonstration that the second order equation ;cyybyu 0a, b, c, constant always has at least one solution of the formxm1 ey 1 , where 1m isa constant.2. Two. Explain why E. D. 1st point must have, consequently, a second solution of theformxm2 ey 2 or formxm2 xey 2 , where 1m y 2m are constants.HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTSWe have seen that the first order linear equation, 0 uydxdy, where a is a constant, has theexponential solutionax1 ecy 1 ranging  ;.-  therefore as natural to try to determine ifthere are exponential solutions  .- homogeneous linear equations of higher order type:   001211   yayayayaya nnnn  (1)Where the coefficients ,1,0,i,ai  n are real constants and 0na . To our surprise, allsolutions of the equation (1) are exponential functions or are formed from exponentialfunctions.
2. 2. Prof. Enrique Mateus NievesPhD in Mathematics Education.2Method of solution: start with the special case of the second order equation ay” + by’ + cy =0. (2) If we try a solution of the formmxey  , thensmxmey  andmxemy 2 , so thatthe equation (2) becomes:, 0 mxmxmx2cebmeeam , or  02 cbmamemx. Asmxe never zero when x has real value, the only way that theexponential function satisfies the differential equation is choosing a m such that it is a root ofthe quadratic equation   02 cbmam (3).This equation is called auxiliary equation or characteristic equation of the differential equation(2). Examine three cases: the solutions of the auxiliary equation corresponding to distinct realroots, real and equal roots and complex conjugate roots.CASE I: distinct real roots:If equation (3) has two distinct real roots, 1m y 2m , arrived at two solutions,xm1 ey 1 andxm2 ey 2 . These functions are linearly independent on  .- and, therefore, form afundamental set. Then, the general solution of equation (2) in this interval isxmxmH ececy 2121  (4)CASE II: Real Estate and equalWhen 21 mm  we necessarily exponential only solution,xm1 ey 1 . According to thequadratic formula,abm12 because the only way 21 mm  is 04  acb2. Thus, a secondsolution of the equation is:  xmxmxmxmxmH xedxeeeey 1111122(5)In this equation we take thatab2m1 . The general solution is thereforexmxmH xececy 1121  (6)
3. 3. Prof. Enrique Mateus NievesPhD in Mathematics Education.3CASE III: complex conjugate roots.If 21 m,,m  are complex, we can write im1   y im2   , where0and  and p > 0 and they are real, and e .i21 There is no formal differencebetween this case and case I, hence,x)i(ececy x)i(H  21 However, inpractice it is preferred to work with real functions and not complex exponential. With this objectusing Eulers formula: ,senicosei that  is a real number. The consequence ofthis formula is that: ,xsenixcose xi and ,xsenixcose x-i (7) where wehave used x)(cosx)(-cos   and x)(senx)(-sen   . Note that if the first add andthen subtract the two equations (7), we obtain respectively:,xcosee x-ixi2 y .xsiniee x-ixi2Asx)i(ececy x)i(   21 is a solution to equation (2) for any choice of theconstants 1c and, 2c if 121  cc and ,c 11  12 c obtain the solutions:x)i(eey x)i(1  yx)i(eey x)i(2  But   xcoseeeey xxixix1 2 and  xsinieeeey xxixix2 2 Accordingly, the results demonstrate that the last two real functions xcose xandxsene xare solutions of the equation (2). Moreover, these solutions form a fundamental,therefore .- , the general solution is: xsincxcoscexsinecxcosecy2xxxH121Second order differential equationsSolve the following differential equations:1. 0352  yyy
4. 4. Prof. Enrique Mateus NievesPhD in Mathematics Education.4SOLUTION: I present the auxiliary equations, roots and corresponding general solutions.   33120352 2112 2m;mmmmm Hence:xececyx32122. 02510  yyySOLUTION:   50502510 122 2mmmmm Hence:xxxececy 5251 3. 0 yyySOLUTION: ,i21-m,i21-mmm 2232301 12 Hence:x23sincx23cosceyxH 2124. Initial value problem. Solve the initial value problem  20134  0y-1,(0)y;yyySOLUTION: The roots of the auxiliary equationi32m,i32mmm 2  120134 so that x3sincx3coscey xH 212By applying the condition 10 )(y , we see that  0senc0cosce 2101  and11 c . We differentiate the above equation and then applying y’(O) = 2 we get232 2  c or342 c ; therefore, the solution is: x3senx3cosey xH342
5. 5. Prof. Enrique Mateus NievesPhD in Mathematics Education.5The two differential equations, 0 pyy y 02  kyy , k real, are important inapplied mathematics. For the first, the auxiliary equation 022 km has imaginary rootsikm1  y ikm1  . According to equation (8), with 0 y k, the general solution iskxsenckxcoscy 2 1 (9)Auxiliary equation the second equation, 022 km , has distinct real roots km1  ykm1  ; therefore its general solution is-kx2kxececy  1 (10)Note that if we choose 212  cc1 and then 21221 cyc1 in  ,10 particular solutions wekxcosheey-kxkx2and kx.senheey-kxkx2inasmuch as kxcosh andkxsenh are linearly independent in any range of the x axis, an alternative form of thegeneral solution of 0 pyy is kxsenhckxcoshcy 2Higher-Order EquationsIn general, to solve a differential equation of order n as   001211   yayayayaya nnnn  (11)Where n,20,1,i,ai  are real constants, we must solve a polynomial equation of degreen:   0012211   amamamama nnnn  (12)If all the roots of equation (12) are real and distinct, the general solution of equation (11) is.ecececy xmnxm2xm n21 1Its difficult to summarize the analogous cases II and III because the roots of an auxiliaryequation of degree greater than two can occur in many combinations. For example, a quinticcould have five distinct real roots, or three distinct real roots and two complex, or four real andcomplex, five reals but equal, but two equals five reals, and so on. When 1m is a root of anequation k multiplicity auxiliary degree n (ie roots equals k), one can show that the solutionsare linearly independent
6. 6. Prof. Enrique Mateus NievesPhD in Mathematics Education.6.exex,xe,e xmkxmxmxm 1111 12 Finally, remember that when the coefficients are real, complex roots of auxiliary equationalways appear in conjugate pairs. Thus, for example, a cubic polynomial equation may havetwo complex zeros at most.Third-order differential equationResolve 043  yyy y”’ + 3~” - 4y = 0.SOLUTION: In reviewing 043 23 mm we should note that one of its roots is 11m .If we divide 43 23 mm eight ,m 1 we see that      ,mmmmmmm 22144143 223 and then the other roots are2 32 mm . Thus, the general solution is.xecececy -2x-2x2x31 Fourth-order differential equationResolve 02 2244 ydxyddxydSOLUTION: The auxiliary equation is   010132224 mmm and has theroots imm 31  y imm 42  . Thus, according to the case II, the solution is:.ecxecececy -ix4ix-ix2ix 31According to Eulers formula, we can write the grouping-ix2ixecec 1 in the formxsencxcosc 21 With a change in the definition of the constants. equally, -ix4ixececx 3 can be expressed in the form  xsencxcoscx 43 Accordingly, thegeneral solution is.x.sinxcxcosxcxsincxcoscy 42  31
7. 7. Prof. Enrique Mateus NievesPhD in Mathematics Education.7General Exercises Nonhomogeneous linear equations with constant coefficientsof order two and higher.For each of the following E. D. finds the general solution:1. 04  yy 2. 052  yy3. 036  yy 4. 08  yy5. 09  yy 6. 03  yy7. 06  yyy 8. 023  yyy9. 016822 ydxdydxyd10. 0251022 ydxdydxyd11. 053  yyy 12. 048  yyy13. 02512  yyy 14. 028  yyy15. 054  yyy 16 0432  yyy17. 023  yyy 18. 022  yyy19. 054  yyy 20. 044  yyy21. 0 yy 22. 05  yy23. 0935  yyyy 24. 01243  yyyy25. 02  yyy 26. 04  yyy27. 09223344 ydxyddxyddxyd28. 051025 22334455 ydxdydxyddxyddxyddxyd